SATHEE: Work Energy And Power Question 230

Work Energy And Power Question 230

Question: A force acts on a 30gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

Options:

A) 576mJ

B) 450mJ

C) 490mJ

D) 530mJ

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $x = 3 t - 4 t^{2} + t^{3} = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

Acceleration $= \frac{d^{2} x}{d t^{2}} = - 8 + 6 t$

Acceleration after $4 s e c = - 8 + 6 \times 4 = 16 m s^{- 2}$

Displacement in $4 s e c = 3 \times 4 - 4 \times 4^{2} + 4^{3} = 12 m$

$∴$ Work = Force $\times$ displacement $= M a s s \times a c c . \times d i s p . = 3 \times 1 0^{- 3} \times 16 \times 12 = 576 m J$

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Work Energy And Power Question 230

Question: A force acts on a 30gm particle in such a way that the position of the particle as a function of time is given by x=3t−4t2+t3 , where x is in metres and t is in seconds. The work done during the first 4 seconds is

Options:

Answer:

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Question: A force acts on a 30gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is